Question

A nucleus at rest undergoes a decay, emitting an $\alpha$ particle of de-Broglie wavelength $\lambda =5.76\times {10}^{-15}\text{m}$. If the mass of the daughter nucleus is $223.610\text{u}$ and that of the $\alpha -$ particle is $4.002\text{u}$, the mass of the parent nucleus in $\text{amu}$ nearly is (integer only)

Take : $1\text{u}=931.5\text{MeV}/c^{2}h=6.62\times {10}^{-34}\text{J s}$

Answer 1

Given, $m_d=223.610\text{u};m_{\alpha }=4.002\text{u}{\lambda }_{\alpha }=5.76\times {10}^{-15}\text{m}$

From de-Broglie wavelength,

${\lambda }_{\alpha }=\frac{h}{P_{\alpha }}$

$\Rightarrow P_{\alpha }=\frac{h}{{\lambda }_{\alpha }}=\frac{6.62\times {10}^{-34}}{5.76\times {10}^{-15}}\approx 1.15\times {10}^{-19}$

By conservation of linear momentum, we have

$0={\overrightarrow{P}}_{\alpha }+{\overrightarrow{P}}_d$

$\Rightarrow P_{\alpha }=-P_d$

$\Rightarrow \left|\begin{array}{c}{\overrightarrow{P}}_{\alpha }\end{array}\right|=\left|\begin{array}{c}{\overrightarrow{P}}_d\end{array}\right|=P$

The kinetic energy released in the process is,

$K=K_{\alpha }+K_d$

$=\frac{P^2}{2m_{\alpha }}+\frac{P^2}{2m_d}$

$=\frac{P^2}{2m_{\alpha }}(1+\frac{m_{\alpha }}{m_d})$

After substituting the given values, we get

$K=6.25\text{MeV}$

If $m_p$ is the mass of the parent nucleus, then

$K+(m_d+m_{\alpha })c^2=m_p{c}^2$

$6.25+(223.61+4.002)c^2=m_p{c}^2$

$\frac{6.25\text{u}\times c^2}{931.5}+(223.61\text{u}+4.002\text{u})c^2=m_p{c}^2$

$m_p=227.62\text{u}\approx 228$